Hypebol \((H):\frac{{{x^2}}}{9} - \frac{{{y^2}}}{7...

D

- Hướng dẫn giải

Phương pháp giải:

\(M({x_0};{y_0}) \in (H) \Rightarrow \left\{ \begin{array}{l}M{F_1} = \left| {a + \frac{{c{x_0}}}{a}} \right|\\M{F_2} = \left| {a - \frac{{c{x_0}}}{a}} \right|\end{array} \right.\)

Giải chi tiết:

\((H):\frac{{{x^2}}}{9} - \frac{{{y^2}}}{7} = 1 \Rightarrow \left\{ \begin{array}{l}a = 3\\b = \sqrt 7 \\c = \sqrt {{a^2} + {b^2}}  = 4\end{array} \right.\)

\(M({x_0};{y_0}) \in (H) \Rightarrow \left\{ \begin{array}{l}M{F_1} = \left| {a + \frac{{c{x_0}}}{a}} \right| = \left| {3 + \frac{4}{3}{x_0}} \right|\\M{F_2} = \left| {a - \frac{{c{x_0}}}{a}} \right| = \left| {3 - \frac{4}{3}{x_0}} \right|\end{array} \right.\)

Mà \(M{F_1} + M{F_2} = 10 \Rightarrow \left| {3 + \frac{4}{3}{x_0}} \right| + \left| {3 - \frac{4}{3}{x_0}} \right| = 10\)

+) TH1: \({x_0} <  - \frac{9}{4} \Rightarrow  - 3 - \frac{4}{3}{x_0} + 3 - \frac{4}{3}{x_0} = 10 \Leftrightarrow  - \frac{8}{3}{x_0} = 10\, \Leftrightarrow {x_0} =  - \frac{{15}}{4}\,\,\,\left( {tm} \right)\)

+)  \( - \frac{9}{4} \le {x_0} \le \frac{9}{4} \Rightarrow 3 + \frac{4}{3}{x_0} + 3 - \frac{4}{3}{x_0} = 10 \Leftrightarrow 6 = 10\,\) (vô lý)

+)  \({x_0} > \frac{9}{4} \Rightarrow 3 + \frac{4}{3}{x_0} - 3 + \frac{4}{3}{x_0} = 10 \Leftrightarrow \frac{8}{3}{x_0} = 10 \Leftrightarrow {x_0} = \frac{{15}}{4}\,\,\,\,\left( {tm} \right)\)

Mà \(M\) thuộc nhánh phải \( \Rightarrow {x_0} > 0 \Rightarrow {x_0} = \frac{{15}}{4}\)

Chọn: D